Solving a Delta Connected Induction Machine Homework Problem

[VinnyCee]

Homework Statement

A \Delta connected induction machine is to start from a source of 480V, 60 Hz.

The motor ratings are :

4 poles, \Delta connected, 480V, 60 Hz, X_{ls}\,=\,X_{lr}\,=\,4\,\Omega, X_m\,=\,50\,\Omega, R_s\,=\,0.25\,\Omega, R_r\,=\,0.4\,\Omega.

Find the current in the windings and the torque of the motor and the current at the line side.

Homework Equations

I_S\,=\,\frac{V_{l-l}}{\left(R_s\,+\,j\,X_{ls}\right)\,+\,\left(j\,X_m\,||\,R_r\,+\,j\,X_{lr}\right)}

I_R\,=\,I_S\,\frac{j\,X_m}{R_r\,+j\,X_{lr}\,+\,j\,X_m}

P_{gap}\,=\,n_{ph}\,|I_R|^2\,\left(\frac{R_r}{s}\right)

Where n_{ph} is number of stator phases.

T\,=\,3\,\frac{P_{gap}}{\omega_s}\,\frac{p}{2}

Where p is number of poles.

The above equations are not given explicitly anywhere, but an example uses them so I think they are right.

The Attempt at a Solution

At starting, s = 1.

I_S\,=\,\frac{480}{\left(0.25\,+\,j\,4\right)\,+\,\left(j\,50\,||\,0.4\,+\,j\,4\right)}\,=\,62.1^{\angle\,-85.6^{\circ}}

I_R\,=\,\left(62.1^{\angle\,-85.6^{\circ}}\right)\,\frac{j\,50}{0.4\,+j\,4\,+\,j\,50}\,=\,57.5^{\angle\,-85.2^{\circ}}

P_{gap}\,=\,(3)\,(57.5)^2\,(0.4)\,=\,3967.5

T\,=\,3\,\left(\frac{3967.5}{377}\right)\,\left(\frac{0.4}{1}\right)\,=\,63.14\,Nm

Does that look right? How do I get the current at the line side?

 

[VinnyCee]

I found an equation that will possibly give line side current...

I_{line}\,=\,\sqrt{3}\,I_{ph}

But what is I_{ph}? Is it I_S or I_R?